This is my first post and hope I'm not doing anything wrong.
I am trying to write a program that find the first value of the vector that reach k-occurrences in it.
For example, given this vector and k=3:
1 1 2 3 4 4 2 2 1 3
I would see 2 as output, because 2 is the first number reaching the 3rd occurrence.
The following code is what I tried to run, but somehow output is not correct.
#include<iostream>
#include<vector>
using namespace std;
int main()
{
vector<int> vettore;
int k;
int a,b,i;
int occ_a;
int occ_b;
cout<< "Write values of vector (number 0 ends the input of values)\n";
int ins;
cin>>ins;
while(ins)
{
vettore.push_back(ins); //Elements insertion
cin>>ins;
}
cout<<"how many occurrences?\n"<<endl;;
cin>>k;
if(k>0)
{
int i=0;
b = vettore[0];
occ_b=0;
while(i< vettore.size())
{
int j=i;
occ_a = 0;
a = vettore[i];
while(occ_a < k && j<vettore.size())
{
if(vettore[j]== a)
{
occ_a++;
vettore.erase(vettore.begin() + j);
}
else
j++;
}
if(b!=a && occ_b < occ_a)
b = a;
i++;
}
cout << b; //b is the value that reached k-occurrences first
}
return 0;
}
Hours have passed but I have not been able to solve it.
Thank you for your help!
Your code is difficult to read because you are declaring variables where they are not used. So their meanings is difficult to understand.
Also there is no need to remove elements from the vector. To find a value that is the first that occurs k-times is not equivalent to to change the vector. They are two different tasks.
I can suggest the following solution shown in the demonstrative program below.
#include <iostream>
#include <vector>
int main()
{
std::vector<int> v = { 1, 1, 2, 3, 4, 4, 2, 2, 1, 3 };
size_t least_last = v.size();
size_t k = 3;
for ( size_t i = 0; i + k <= least_last; i++ )
{
size_t count = 1;
size_t j = i;
while ( count < k && ++j < least_last )
{
if ( v[j] == v[i] ) ++count;
}
if ( count == k )
{
least_last = j;
}
}
if ( least_last != v.size() ) std::cout << v[least_last] << '\n';
return 0;
}.
The program output is
2
The idea is to find the last position of the first element that occurs k-times. As soon as it is found the upper limit of the traversed sequence is set to this value. So if there is another element that occurs k-times before this limit then it means that it occurs the first compared with already found element.
Related
I am a C++ student. And I need to solve this problem: "Write a program that receives a number and an array of the size of the given number. The program must find all the duplicates of the given numbers, push-back them to a vector of repeating elements, and print the vector". The requirements are I'm only allowed to use the vector library and every repeating element of the array must be pushed to the vector only once, e.g. my array is "1, 2, 1, 2, 3, 4...", the vector must be "1 ,2".
Here's what I've done so far. My code works, but I'm unable to make it add the same duplicate to the vector of repeating elements only once.
#include <iostream>
#include <vector>
int main() {
int n;
std::cin >> n;
int* arr = new int[n];
std::vector<int> repeatedElements;
for(int i = 0; i < n; ++i) {
std::cin >> arr[i];
}
for(int i = 0; i < n; ++i) {
bool foundInRepeated = false;
for(int j = 0; j < repeatedElements.size(); ++j) {
if(arr[i] == repeatedElements[j]) {
foundInRepeated = true;
break;
}
}
if(foundInRepeated) {
continue;
} else {
for(int i = 0; i < n; ++i) {
int count = 1;
for(int j = i + 1; j < n; ++j) {
if(arr[i] == arr[j]) {
++count;
}
}
if(count > 1) {
repeatedElements.push_back(arr[i]);
}
}
}
}
for(int i = 0; i < repeatedElements.size(); ++i) {
std::cout << repeatedElements[i] << " ";
}
std::cout << std::endl;
}
Consider what you're doing here:
if(foundInRepeated) {
continue;
} else {
for(int i = 0; i < n; ++i) { // why?
If the element at some index i (from the outer loop) is not found in repeatedElements, you're again iterating through the entire array, and adding elements that are repeated. But you already have an i that you're interested in, and hasn't been added to the repeatedElements. You only need to iterate through j in the else branch.
Removing the line marked why? (and the closing brace), will solve the problem. Here's a demo.
It's always good to follow a plan. Divide the bigger problem into a sequence of smaller problems is a good start. While this often does not yield an optimal solution, at least it yields a solution, which is more or less straightforward. And which subsequently can be optimized, if need be.
How to find out, if a number in the sequence has duplicates?
We could brute force this:
is_duplicate i = arr[i+1..arr.size() - 1] contains arr[i]
and then write ourselves a helper function like
bool range_contains(std::vector<int>::const_iterator first,
std::vector<int>::const_iterator last, int value) {
// ...
}
and use it in a simple
for (auto iter = arr.cbegin(); iter != arr.cend(); ++iter) {
if (range_contains(iter+1, arr.cend(), *iter) && !duplicates.contains(*iter)) {
duplicates.push_back(*iter);
}
}
But this would be - if I am not mistaken - some O(N^2) solution.
As we know, sorting is O(N log(N)) and if we sort our array first, we will
have all duplicates right next to each other. Then, we can iterate over the sorted array once (O(N)) and we are still cheaper than O(N^2). (O(N log(N)) + O(N) is still O(N log(N))).
1 2 1 2 3 4 => sort => 1 1 2 2 3 4
Eventually, while using what we have at our disposal, this could yield to a program like this:
#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>
using IntVec = std::vector<int>;
int main(int argc, const char *argv[]) {
IntVec arr; // aka: input array
IntVec duplicates;
size_t n = 0;
std::cin >> n;
// Read n integers from std::cin
std::generate_n(std::back_inserter(arr), n,
[](){
return *(std::istream_iterator<int>(std::cin));
});
// sort the array (in ascending order).
std::sort(arr.begin(), arr.end()); // O(N*logN)
auto current = arr.cbegin();
while(current != arr.cend()) {
// std::adjacent_find() finds the next location in arr, where 2 neighbors have the same value.
current = std::adjacent_find(current,arr.cend());
if( current != arr.cend()) {
duplicates.push_back(*current);
// skip all duplicates here
for( ; current != (arr.cend() - 1) && (*current == *(current+1)); current++) {
}
}
}
// print the duplicates to std::cout
std::copy(duplicates.cbegin(), duplicates.cend(),
std::ostream_iterator<int>(std::cout, " "));
return 0;
}
I'm doing a fairly easy HackerRank test which asks the user to write a function which returns the minimum number of swaps needed to sort an unordered vector in ascending order, e.g.
Start: 1, 2, 5, 4, 3
End: 1, 2, 3, 4, 5
Minimum number of swaps: 1
I've written a function which works on 13/14 test cases, but is too slow for the final case.
#include<iostream>
#include<vector>
using namespace std;
int mimumumSwaps(vector<int> arr) {
int p = 0; // Represents the (index + 1) of arr, e.g. 1, 2, ..., arr.size() + 1
int swaps = 0;
for (vector<int>::iterator i = arr.begin(); i != arr.end(); ++i) {
p++;
if (*i == p) // Element is in the correct place
continue;
else{ // Iterate through the rest of arr until the correct element is found
for (vector<int>::iterator j = arr.begin() + p - 1; j != arr.end(); ++j) {
if (*j == p) {
// Swap the elements
double temp = *j;
*j = *i;
*i = temp;
swaps++;
break;
}
}
}
}
return swaps;
}
int main()
{
vector<int> arr = { 1, 2, 5, 4, 3 };
cout << mimumumSwaps(arr);
}
How would I speed this up further?
Are there any functions I could import which could speed up processes for me?
Is there a way to do this without actually swapping any elements and simply working out the min. swaps which I imagine would speed up the process time?
All permutations can be broken down into cyclic subsets. Find said subsets.
Rotating a subset of K elements by 1 takes K-1 swaps.
Walk array until you find an element out of place. Walk that cycle until it completes. Advance, skipping elements that you've put into a cycle already. Sum (size-1) for each cycle.
To skip, maintain an ordered or unordered set of unexamined items, and fast remove as you examine them.
I think that gives optimal swap count in O(n lg n) or so.
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
using namespace std;
int minimumSwaps(vector<int> arr)
{
int i,c,j,k,l;
j=c=0;
l=k=arr.size();
while (j<k)
{
i=0;
while (i<l)
{
if (arr[i]!=i+1)
{
swap(arr[i],arr[arr[i]-1]);
c++;
}
i++;
}
k=k/2;
j++;
}
return c;
}
int main()
{
int n,q;
cin >> n;
vector<int> arr;
for (int i = 0; i < n; i++)
{
cin>>q;
arr.push_back(q);
}
int res = minimumSwaps(arr);
cout << res << "\n";
return 0;
}
Can someone help me with this excercise?
Describe an algorithm in C/C++ which:
Defines 2 vectors: first named a which contains 1000 integers and second one b which contains 500 integers
edit vector a by inserting in every position a value read from standard input. It is assumed that a maximum of 500 values are positive numbers.
after point 2, edit vector b inserting many 1 as many positive numbers are there in the vector a. Remaining portions of vector b must contain the value 0.
This:
#include<cstdlib>
#include <iostream>
using namespace std;
int main() {
int A[1000], B[1000]; int i;
for(i=0;i<1000;i++) {
cout<<"Initialising vector A: ";
cin>>A[i];
}
int j;
for(i=0,j=999;i<1000;i++,j--)
B[j]=A[i];
cout<<"Vector B is: ";
for(i=0;i<1000;i++)
cout<<B[i]<<"\t";
system("pause");
return 0;
}
is a similar excercise I did with different requests, but now I don't know how to edit it to fit the new requirements.
All you need is a loop like this
int B[500] = {};
//...
int n = 0;
for ( int i = 0, i < 1000 ; i++ )
{
if ( A[i] > 0 ) B[n++] = 1;
}
for ( int i = 0; i < n; i++ ) cout << B[i] << ' ';
cout << endl;
#include <iostream>
using namespace std;
class A
{
public:
int index;
int t;
int d;
int sum;
};
A arr[1000];
bool comp (const A &a, const A &b)
{
if (a.sum < b.sum)
return true;
else if (a.sum == b.sum && a.index < b.index)
return true;
return false;
}
int main (void)
{
int n,foo,bar,i;
i = 0;
cin>>n;
while ( n != 0 )
{
cin>>foo>>bar;
arr[i].index = i+1;
arr[i].t = foo;
arr[i].d = bar;
arr[i].sum = arr[i].t+arr[i].d;
n--;
i++;
}
sort(arr,arr+n,comp);
for ( int j = 0; j < n; j++ )
cout<<arr[j].index;
cout<<"\n";
return 0;
}
So, I made this program which accepts values from users, and then sorts them on the basis of a specific condition. However, when I try to print the values of the array, it doesn't print it. I don't know why. Please help. Thanks!
PS: I even tried to print the value of the array without use of comparator function, but still it doesn't print.
Edit: Got my error, as mentioned by some amazing people in the comments. However, on inputting the values as,
5
8 1
4 2
5 6
3 1
4 3
It should return the answer as 4 2 5 1 3 but instead it returns 1 2 3 4 5. Basically, it sorts according to the sum of the 2 elements in each row and prints the index. What might be the problem?
It is not sorting because you are modifying the value of n in the while loop.
So, when n = 0 after the loop ends, there is nothing to sort (or to print later for that matter):
sort( arr, arr + n, comp);
is equivalent to
sort( arr, arr + 0, comp);
Easiest approach would be to use a for loop:
for( int i=0; i<n; ++i )
{
....
}
This way, n remains unchanged.
With the help of SO members, the following program successfully converts a static 1D array into a 2D vector by considering below criteria:
Each time an element with value = 0 is encountered, a new row is created. Basically when a 0 is encountered, row value is increased and column value is reset to 0. If a non-zero value is encountered, the row value is maintained and column value is increased.
// declarations
int givenArray[9] = {1, 2, 3, 0, 4, 0, 1, 2, 1};
std::vector<int>::size_type j;
std::vector<int>::size_type i;
vector<vector<int>> my2dArray;
vector<int> dArray;
void calc(vector<int>&, int);
int task;
int sum = 0;
int main() {
for (int i = 0; i < 9;
i++) // iterate through all elements of the given array
{
if (i == 0) // adding the first element
{
my2dArray.resize(my2dArray.size() + 1);
my2dArray.back().push_back(givenArray[i]);
continue;
}
if (givenArray[i] == 0) // re-size if 0 is encountered
{
my2dArray.resize(my2dArray.size() + 1);
}
my2dArray.back().push_back(givenArray[i]);
}
for (std::vector<std::vector<int>>::size_type i = 0; i < my2dArray.size();
i++) {
for (std::vector<int>::size_type j = 0; j < my2dArray[i].size(); j++) {
std::cout << my2dArray[i][j] << ' ';
if (my2dArray[i].size() > 2) {
task = my2dArray[i].size();
calc(my2dArray[i], task);
}
}
std::cout << std::endl;
}
}
void calc(vector<int>& dArray, int task) {
int max = 0;
for (unsigned int j = 0; j < task; j++) {
if (dArray[i] > max)
dArray[i] = max;
}
cout << "\nMax is" << max;
}
However, I want to pass a single row of 2D vector 2dArray to function calc if the number of columns for each row exceeds 2. Function calc aims to find maximum value of all the elements in the passed row. The above program doesn't yield the desired output.
Some improvements:
i and j global variables are not needed, you are declaring the variables of the loops in the loop initialization (ex: for (int i = 0; i < 9; i++), the same for the other loops).
It's better not to used global variables, only when strictly necessary (with careful analysis of why). In this case it's not necessary.
The typedef are for more easy access to inner typedef of the type (ex: size_type).
You were doing the call to calc method in every iteration of the inner loop, and iterating over the same row multiple times, this call should be executed once per row.
Using the size of array givenArray as constant in the code is not recommended, later you could add some elements to the array and forgot to update that constant, it's better to declare a variable and calculated generally (with sizeof).
There is no need to pass the size of the vector to method calc if you are passing the vector.
As recommended earlier it's better to use std::max_element of algorithm header.
If you could use C++11 the givenArray could be converted to an std::vector<int> and maintain the easy initialization.
Code (Tested in GCC 4.9.0)
#include <vector>
#include <iostream>
using namespace std;
typedef std::vector<int> list_t;
typedef std::vector<list_t> list2d_t;
void calc(list_t& dArray, long& actual_max) {
for (unsigned int j = 0; j < dArray.size(); j++) {
if (dArray[j] > actual_max) {
actual_max = dArray[j];
}
}
cout << "Max is " << actual_max << "\n";
}
void calc(list_t& dArray) {
long actual_max = 0;
for (unsigned int j = 0; j < dArray.size(); j++) {
if (dArray[j] > actual_max) {
actual_max = dArray[j];
}
}
cout << "Max is " << actual_max << "\n";
}
int main() {
int givenArray[9] = {1, 2, 3, 0, 4, 0, 1, 2, 1};
int givenArraySize = sizeof(givenArray) / sizeof(givenArray[0]);
list2d_t my2dArray(1);
list_t dArray;
for (int i = 0; i < givenArraySize; i++) {
if (givenArray[i] == 0) {
my2dArray.push_back(list_t());
} else {
my2dArray.back().push_back(givenArray[i]);
}
}
long max = 0;
for (list2d_t::size_type i = 0; i < my2dArray.size(); i++) {
for (list_t::size_type j = 0; j < my2dArray[i].size(); j++) {
std::cout << my2dArray[i][j] << ' ';
}
std::cout << "\n";
if (my2dArray[i].size() > 2) {
// if you need the max of all the elements in rows with size > 2 uncoment bellow and comment other call
// calc(my2dArray[i], max);
calc(my2dArray[i]);
}
}
}
Obtained Output:
1 2 3
Max is 3
4
1 2 1
Max is 2
You have a few problems:
You don't need to loop over j in the main function - your calc function already does this.
Your calc function loops over j, but uses the global variable i when accessing the array.
Your calc function assigns the current max value to the array, rather than assigning the array value to max
Function calc aims to find maximum value of all the elements in the passed row. The above program doesn't yield the desired output.
Instead of writing a function, you could have used std::max_element.
#include <algorithm>
//...
int maxVal = *std::max_element(my2dArray[i].begin(), my2dArray[i].begin() + task);
cout << "\Max is " << maxVal;